# Determine magnitude of force acting on block on inclined plane

In summary, a 20 kg package is at rest on an incline. A force is applied and determined. The static and kinetic coefficients of friction between the package and the incline are both equal to 0.3. If I know the acceleration, I can find the magnitude of P.
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## Homework Statement

A 20-kg package is at rest on an incline when a force P is applied to it. Determine the magnitude of P if 10s is required for the package to travel 5m up the incline. The static and kinetic coefficients of friction between the package and the incline are both equal to 0.3.

Answer: 419N to start and 301 during sliding

## Homework Equations

m=20kg
g=9.8m/s2
t=10s
xi = 0ft
xf = 5ft
μs = μk = 0.3
Frictional force = Ff = μN

## The Attempt at a Solution

First, I drew a free-body diagram including tilted coordinate axes:

Then, I went about finding the sum of the forces equations in the x- and y-directions.

$$\sum F_{x}=-F_{f}-mgsin(20)+Pcos(30)=ma$$

$$\sum F_{y}=N-mgcos(20)-Psin(30)=0$$

From the second equation,
$$N=mgcos(20)+Psin(30)$$

Plugging this value for N into Ff = μN in the first equation yields:

$$P=\frac{m\left(a+\mu gcos(20)+gsin(20)\right)}{cos(30)-\mu sin(30)}$$

If I know the acceleration, I can find the magnitude of P. There has to be a way to find acceleration from the initial conditions, but I'm at a loss. Also, once static friction is overcome, I'm sure the sum of the forces equations will be different but I'm not sure how.

Well since the force seems to be constant, you could apply the kinematic equation to solve for the acceleration:
$$x = x_0 + v_0t + (1/2)at^2$$

Thefox14 said:
Well since the force seems to be constant, you could apply the kinematic equation to solve for the acceleration:
$$x = x_0 + v_0t + (1/2)at^2$$

From the $\sum F_{x}$ equation, along with $F_{f}=\mu N$:

$$a=-\mu gcos(20)-\frac{\mu Psin(30)}{m}-gsin(20)+Pcos(30)$$

$$x_{f}=x_{i}+v_{i}t+\frac{1}{2}at^{2}$$

$$5=\frac{1}{2}\left (-\mu gcos(20)-\frac{\mu Psin(30)}{m}-gsin(20)+Pcos(30) \right )\left (10^{2} \right )$$

$$\frac{1}{10}=-\mu gcos(20)-\frac{\mu Psin(30)}{m}-gsin(20)+Pcos(30)$$

$$\frac{1}{10}+\mu gcos(20)+gsin(20)=P\left (cos(30)-\frac{\mu sin(30)}{m} \right )$$

$$P=\frac{\frac{1}{10}+\mu gcos(20)+gsin(20)}{cos(30)-\frac{\mu sin(30)}{m}}=\frac{\frac{1}{10}+(0.3)(9.8)cos(20)+(9.8)sin(20)}{cos(30)-\frac{0.3sin(30)}{20}}\approx 7.24N \neq \: Book's\: answer\: of\: 419N$$

Did I do the math wrong? Where do I go from here?

## 1. What is the formula for determining the magnitude of force acting on a block on an inclined plane?

The formula for determining the magnitude of force acting on a block on an inclined plane is F = mg sin θ, where F is the force, m is the mass of the block, g is the acceleration due to gravity, and θ is the angle of the inclined plane.

## 2. How does the angle of the inclined plane affect the magnitude of force acting on the block?

The angle of the inclined plane directly affects the magnitude of force acting on the block. As the angle increases, the force required to keep the block from sliding down the plane also increases. This is because a steeper angle increases the component of the force of gravity that acts parallel to the plane.

## 3. How does the mass of the block affect the magnitude of force acting on it on an inclined plane?

The mass of the block also affects the magnitude of force acting on it on an inclined plane. A heavier block will require a greater force to keep it from sliding down the plane, as the force of gravity acting on the block will be greater.

## 4. What is the relationship between the magnitude of force and the angle of the inclined plane?

The magnitude of force and the angle of the inclined plane have a direct relationship. As the angle of the inclined plane increases, the magnitude of force required to keep the block from sliding down the plane also increases.

## 5. How can the magnitude of force on a block on an inclined plane be decreased?

The magnitude of force on a block on an inclined plane can be decreased by decreasing the angle of the inclined plane or by decreasing the mass of the block. Additionally, using frictional forces or adding a weight or counterweight to the block can also decrease the magnitude of force acting on it.

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